The one-to-one functions \( g \) and \( h \) are defined as follows. \begin{tabular}{c}\( g=\{(0,6),(2,3),(3,1),(6,-2),(9,5)\} \) \\ \( h(x)=3 x+13 \) \\ Find the following. \\ \hline\( g^{-1}(3)=\square \) \\ \( h^{-1}(x)=\square \) \\ \( \left(h^{-1} \circ h\right)(-7)=\square \)\end{tabular}

To find \( g^{-1}(3) \), we need to look for an ordered pair in the set \( g \) where the second element is 3. Scanning through the pairs, we find that \( (2, 3) \) is the relevant pair. Thus, \( g^{-1}(3) = 2 \). Next, we calculate \( h^{-1}(x) \). Since \( h(x) = 3x + 13 \), we can find the inverse by solving for \( x \): \[ y = 3x + 13 \implies y - 13 = 3x \implies x = \frac{y - 13}{3} \] Thus, \( h^{-1}(x) = \frac{x - 13}{3} \). Finally, we compute \( (h^{-1} \circ h)(-7) \): First, find \( h(-7) \): \[ h(-7) = 3(-7) + 13 = -21 + 13 = -8 \] Now, use this result in \( h^{-1} \): \[ h^{-1}(-8) = \frac{-8 - 13}{3} = \frac{-21}{3} = -7 \] Therefore, \( (h^{-1} \circ h)(-7) = -7 \). Bringing it all together: \[ g^{-1}(3) = 2, \quad h^{-1}(x) = \frac{x-13}{3}, \quad (h^{-1} \circ h)(-7) = -7 \]